A chemist dissolves 781mg of pure hydroiodic acid in enough water to make up 281mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.

The pH of the solution is 1.66

Explanation:

The formula for hydroiodic acid is HI where the atomic mass of hydrogen (H) is 1 g/mol and that of iodine (I) is 127 g/mol.

The formula for pH is

pH = -log[H⁺]

To determine the hydrogen ion concentration, [H⁺], we calculate the molarity of the solution.

The molarity (M) of the solution is

M = number of moles of solute/volume (in dm³ or L)

volume of solution according to the question is 281 mL ⇔ 0.281 L

number of moles of solute (n) = given mass/molar mass

molar mass of HI = 1 + 127 = 128 g

given mass = 781 mg ⇔ 0.781

n = 0.781/128

n = 0.00610

Molarity = 0.00610/0.281

M = 0.0217M

pH = -log0.0217

pH = 1.66

The pH of the solution is 1.66

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